A uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is _______.
(g is acceleration due to gravity)
Step 1: Understanding the Concept:
The system behaves as a physical pendulum because it is an extended rigid body oscillating about an offset pivot axis. The time period depends on the moment of inertia about the pivot and the distance from the pivot to the center of mass.
Step 2: Key Formula or Approach:
Time period of a physical pendulum: $T = 2\pi \sqrt{\frac{I_{pivot}}{Mgd}}$
where $I_{pivot}$ is the moment of inertia about the pivot, and $d$ is the distance from the pivot to the center of mass.
Parallel Axis Theorem: $I_{pivot} = I_{CM} + Md^2$.
Step 3: Detailed Explanation:
For a uniform disc, the center of mass (CM) is exactly at its geometric center.
The axis A is on the circumference of the disc. Therefore, the distance $d$ from the pivot to the CM is equal to the radius $R$.
$d = R$.
The moment of inertia of a uniform disc about an axis passing through its CM perpendicular to its plane is:
$I_{CM} = \frac{1}{2}MR^2$.
Using the Parallel Axis Theorem, calculate the moment of inertia about the pivot axis A (which is parallel to the central axis and at distance R):
$I_{pivot} = I_{CM} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Now plug these parameters into the physical pendulum formula:
$T = 2\pi \sqrt{\frac{I_{pivot}}{Mgd}}$
$T = 2\pi \sqrt{\frac{\frac{3}{2}MR^2}{MgR}}$
Cancel out $M$ and one factor of $R$:
$T = 2\pi \sqrt{\frac{3R}{2g}}$.
Step 4: Final Answer:
The time period is $2\pi \sqrt{\frac{3R}{2g}}$.
